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Suppose we have a graph embedded on a surface $Q$ and one face $F$ of the graph is not homeomorphic to an open disk. Does there exist a closed (smooth nonselfinteresecting) curve $g$ contained in $F$ such that $g$ does not divide $Q$ into two regions?

Can I use this to prove Youngs Theorem that in any minimal genus embedding all faces are homeomorphic to open disks?

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  • $\begingroup$ So a face is a connected component of the component of the graph $G$? $\endgroup$ Commented May 17 at 8:30
  • $\begingroup$ @ArcticChar yes a face is a connected component $C$ of $Q-G$ I'll also refer to the nodes of $G$ bounding $C$ as the face. $\endgroup$
    – Hao S
    Commented May 17 at 17:01
  • $\begingroup$ There is still very little context , and I am surprised that this has been reopened. $\endgroup$
    – Peter
    Commented May 26 at 6:14
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    $\begingroup$ @Peter first most questions on here have very little context. Second I explained exactly why I'm interested in this question. $\endgroup$
    – Hao S
    Commented May 26 at 6:26
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    $\begingroup$ @Peter What do you mean by context here? $\endgroup$
    – Hao S
    Commented May 26 at 6:43

2 Answers 2

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I'll assume the graph, denoted $G$, is connected, and that the surface $Q$ is closed, connected, and oriented; I believe those are the prerequisites for the definition of genus.

It is not true that such a curve exists. For example, in the torus $T=S^1 \times S^1$ consider the circle graph $G = S^1 \times \{\text{1 point}\} \subset T$. The complement $T-G$ is an open annulus, homeomorphic to $S^1 \times \{\text{an open interval}\}$, and every simple closed curve in an open annulus separates the annulus into two components.

I would suggest a different way to prove that theorem about minimal genus embeddings. If some component of $Q-G$ is not homeomorphic to an open disc then there exists a simple closed curve $C \subset Q-G$ such that $C$ is homotopic to a closed curve in $G$ (possibly not simple), but $C$ is not homotopic in $Q-G$ to a point. By doing surgery along $C$ --- cutting $Q$ along $C$ and gluing in two discs --- you get a smaller genus embedding of $G$.

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    $\begingroup$ I said that the curve does not separate the surface $Q$ into two connected components not that it doesn't separate $Q \backslash G$ into two separate components. So in this case it means the curve doesn't separate the annulus into two separate components. $\endgroup$
    – Hao S
    Commented May 4 at 18:34
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    $\begingroup$ You can show $G$ exists using the classification of surfaces. $\endgroup$
    – Lee Mosher
    Commented May 4 at 18:59
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    $\begingroup$ Yes, I meant $C$. $\endgroup$
    – Lee Mosher
    Commented May 4 at 19:47
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    $\begingroup$ Every compact surface with boundary is homeomorphic to a closed surface minus the interiors of some finitely many pairwise disjoint embedded closed discs. $\endgroup$
    – Lee Mosher
    Commented May 4 at 20:08
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    $\begingroup$ Here is an MSE link on the subject. $\endgroup$
    – Lee Mosher
    Commented May 4 at 20:10
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Does there exist a closed (smooth nonselfinteresecting) curve $g$ contained in $F$ such that $g$ does not divide $Q$ into two regions?

Yes.

Can I use this to prove Youngs Theorem that in any minimal genus embedding all faces are homeomorphic to open disks?

Yes.


See the following paper where exactly this is discussed:

As Lee Mosher suggests in his answer, one would like to perform surgery along an appropriate noncontractible curve in a face $F$. However, a naive approach to this will not always work, as I mentioned in a comment. In the above paper, the authors instead use Theorem 7.2 due to Morton Brown (Locally flat imbeddings of topological manifolds, Ann. Math. (2) 75, 331–341 (1962), Zbl 0201.56202) and the "scissors theorem" (Theorem 2.3) that describes how one can cut a surface along a graph and paste it back together, in order to prove Youngs's theorem (Theorem 7.3).

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